---
title: "Classical: node voltage va(t) with 3 H"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-va-15v-10ohm-3h"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-va-15v-10ohm-3h
site: BCT Engineering
---

# Classical: node voltage va(t) with 3 H

Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution

- Asked in: 2071 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-va-15v-10ohm-3h
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

Find va(t) for t > 0 in the figure below using classical method.
[Figure: 15 V source feeding a top wire with a 10 Ω resistor from the source positive node to the top-right node (bridging), and a second path with a 10 Ω resistor to node va and a 20 Ω resistor from va to the top-right node. A 3 H inductor connects the top-right node to the bottom wire. From va, a switch (t = 0) in series with a 10 Ω resistor connects to the bottom wire.]

## Short answer

Before switch, va=10 V, iL=2 A. Closing the switch gives iL(t)=1.8+0.2e^{-2.381t} A, hence va(t)=6−0.2857e^{-2.381t} V for t>0.

## Answer

**Assumptions from the figure:** 15 V source with + terminal at node S. A 10 Ω resistor connects S to the right node R. A second path: 10 Ω from S to node va, then 20 Ω from va to R. The 3 H inductor connects R to ground. A switch in series with 10 Ω connects va to ground; the switch is open until t = 0 and closes at t = 0.

#### Before t = 0 (switch open, dc steady state)

The inductor is a short (V_R = 0):

```text
Path S–10–va–20–R: current = 15/30 = 0.5 A;  va(0⁻) = 15 − 5 = 10 V
Direct 10 Ω: 15/10 = 1.5 A
i_L(0⁻) = 1.5 + 0.5 = 2 A = i_L(0⁺)
```

#### After t = 0 (switch closed)

Write the node equations with the inductor current i_L (flowing from R to ground):

```text
Node va:  (15 − va)/10 = va/10 + (va − V_R)/20   →   va = 6 + V_R/5
Node R:   (15 − V_R)/10 + (va − V_R)/20 = i_L
          2(15 − V_R) + (va − V_R) = 20 i_L
          36 − 2.8 V_R = 20 i_L      →   V_R = (36 − 20 i_L)/2.8
```

Inductor equation V_R = 3 di_L/dt:

```text
di_L/dt = (36 − 20 i_L)/8.4
di_L/dt + 2.381 i_L = 4.286
```

Time constant τ = 1/2.381 = 0.42 s (equivalent to L/R_th = 3/7.143). Final value i_L(∞) = 36/20 = 1.8 A.

```text
i_L(t) = 1.8 + (2 − 1.8) e^(−2.381t) = 1.8 + 0.2 e^(−2.381t) A
V_R = 3 di_L/dt = −1.4286 e^(−2.381t) V
```

#### Result for va

```text
va(t) = 6 + V_R/5 = 6 − 0.2857 e^(−2.381t) V,   t > 0
```

Check: va(0⁺) = 5.714 V (while va(0⁻) = 10 V, so it jumps because the switch closes). At t → ∞ va = 6 V, since V_R → 0 (inductor shorted).

**Answer: va(t) = 6 − 0.2857e^(−2.381t) V for t > 0.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is i_L(0-) with the switch open?
   A. 2 A: 1.5 A in the direct 10 ohm plus 0.5 A through 10 + 20
   B. 1.5 A
   C. 0.5 A
   D. 3 A
   Answer: A. The inductor is a short so both paths connect source to ground and their currents add. (Key point: Initial state)

2. What is va(0-)?
   A. 15 V
   B. 10 V
   C. 5 V
   D. 6 V
   Answer: B. 15 - 10 ohm x 0.5 A = 10 V. (Key point: Va(0-))

3. What is the equation for i_L after closing the switch?
   A. di_L/dt + 3 i_L = 15
   B. di_L/dt + 7.143 i_L = 8
   C. di_L/dt + 2.381 i_L = 4.286
   D. di_L/dt + 0.42 i_L = 2
   Answer: C. Node equations give V_R = (36 - 20 i_L)/2.8 and V_R = 3 di/dt. (Key point: Differential equation)

4. What is va(infinity)?
   A. 10 V
   B. 15 V
   C. 0 V
   D. 6 V, since V_R tends to 0 and va = 6 + V_R/5
   Answer: D. At dc the inductor is shorted so V_R = 0. (Key point: Final va)

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